11
$\begingroup$

I'm working on some exercise regarding the spin coupling of two electrons. There we have the wavefunctions corresponding to the S values as

$$\begin{align} S = 1: &\begin{array}{c}\uparrow\uparrow \\ \dfrac{1}{\sqrt 2}(\uparrow\downarrow+\downarrow \uparrow) \\ \downarrow\downarrow\end{array} \\[5mm] S = 0: &\ \frac{1}{\sqrt{2}} (\uparrow\downarrow-\downarrow\uparrow) \end{align}$$

I understand that the three wavefunctions for $S=1$ are symmetric and the one for $S=0$ is antisymmetric. My question is, why is the combination with the minus sign is the one for $S=0$?

My thinking is that in a $\uparrow \downarrow$ or $\downarrow\uparrow$ combination the $S_z$ components would already add up to $0$ so that the minus sign would not change anything in the fact that $S=0$.

Or is it that $\downarrow\uparrow$ or $\uparrow\downarrow$ each represent a state with $S=1$ and $S_z = 0$ so that subtracting the one from the other gives $S=1-1=0$?

$\endgroup$
  • $\begingroup$ Just to point out that you can't subtract total spins like $S=1-1$, $S$ represents (semi-classically) the magnitude of a vector and they can't just be added like this. $\endgroup$ – Quantum spaghettification Nov 2 '16 at 21:49
  • $\begingroup$ Ok but how do you explain that $S=0$ for the antisymmetric wavefunction and $S=1$ for the symmetric one? $\endgroup$ – Suppenkasper Nov 2 '16 at 21:58
  • 1
    $\begingroup$ I changed your coefficients to $\frac{1}{\sqrt{2}}$; hopefully that's correct. $\endgroup$ – David Z Nov 2 '16 at 22:08
10
$\begingroup$

You are correct that one of the $S = 1$ states has zero angular momentum along the $z$ axis. However, the $S = 0$ state has zero angular momentum along any direction, and this follows directly from the presence of the minus sign.

To see why, consider the operator for spin along a general direction, $S_{\hat{n}}$. Since the spins are identical, it doesn't matter which spin is what, so $$\langle \uparrow \downarrow | S_{\hat{n}} | \uparrow \downarrow \rangle = \langle \downarrow \uparrow | S_{\hat{n}} | \downarrow \uparrow \rangle = \alpha(\hat{n}).$$ Moreover, the operator $P$ that interchanges the spins doesn't affect $S_{\hat{n}}$, because $$P S_{\hat{n}} = P \left(S_{\hat{n}}^1 + S_{\hat{n}}^2 \right) = S_{\hat{n}}^2 + S_{\hat{n}}^1 = S_{\hat{n}}$$ which further implies that $$\langle \uparrow \downarrow | S_{\hat{n}} | \downarrow \uparrow \rangle = \langle \downarrow \uparrow | S_{\hat{n}} | \uparrow \downarrow \rangle = \alpha(\hat{n}).$$ This is all the information we need to evaluate the spin along the $\hat{n}$ direction for the two states you give. For the $S = 1$ state, we find $$\frac{\alpha}{2} + \frac{\alpha}{2} + \frac{\alpha}{2} + \frac{\alpha}{2} = 2\alpha$$ while for the $S = 0$ state we find $$\frac{\alpha}{2} + \frac{\alpha}{2} - \frac{\alpha}{2} - \frac{\alpha}{2} = 0.$$ The cross terms pick up a minus sign, due to the minus sign in the superposition. This shows the singlet has zero spin along any direction.

$\endgroup$
4
$\begingroup$

It's important to remember that when we say $S=1, 0$ we're really thinking of the eigenstates and eigenvalues of $\mathbf{S}^2 = (\mathbf{S}_1 + \mathbf{S}_2)^2$. Try applying this operator, and you'll see that the symmetric states have eigenvalue $S=1$, and the antisymmetric will have eigenvalue $S=0$. If you're simply applying $\mathbf{S}_z$, which is what you are doing in your question statement, then definitely both the mixed symmetric and antisymmetric states have eigenvalue $j=0$, as you've pointed out.

$\endgroup$
1
$\begingroup$

Just a slightly different view using the interchange operator $\hat P|m_1m_2 \rangle $ = $|m_1m_2 \rangle $

My question is, why is the combination with the minus sign is S=0?

Another way to look at this is:

$$\begin{align} S = 1: &\begin{array}{c}\uparrow\uparrow \\ \dfrac{1}{\sqrt 2}(\uparrow\downarrow+\downarrow \uparrow) \\ \downarrow\downarrow\end{array} \\[5mm] S = 0: &\ \frac{1}{\sqrt{2}} (\uparrow\downarrow-\downarrow\uparrow) \end{align}$$

The triplet states are not affected by the interchange of $ m_1\iff m_2$ whereas the $S=0$ changes sign under interchange.

Using $\hat P|m_1m_2 \rangle $ = $|m_1m_2 \rangle $ as the interchange operator.

This results in

$$\hat P \textrm{(triplet)} = \textrm{triplet} $$ But $$\hat P\textrm{(singlet)} = -~\textrm{ singlet} $$

$\endgroup$
  • $\begingroup$ Do you mean $\hat P|m_1m_2 \rangle = |m_2m_1 \rangle$? $\endgroup$ – Wood Nov 10 '17 at 11:36

protected by Qmechanic Nov 2 '16 at 22:49

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.